poj 2446 Chessboard (二分匹配)

Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12800		Accepted: 4000

Description

Alice and Bob often play games on chessboard. One day, Alice draws a board with size M * N. She wants Bob to use a lot of cards with size 1 * 2 to cover the board. However, she thinks it too easy to bob, so she makes some holes on the board (as shown in the figure below). 

We call a grid, which doesn’t contain a hole, a normal grid. Bob has to follow the rules below: 
1. Any normal grid should be covered with exactly one card. 
2. One card should cover exactly 2 normal adjacent grids.

Some examples are given in the figures below: 
 
A VALID solution.
 
An invalid solution, because the hole of red color is covered with a card.
 
An invalid solution, because there exists a grid, which is not covered.
Your task is to help Bob to decide whether or not the chessboard can be covered according to the rules above.

Input

There are 3 integers in the first line: m, n, k (0 < m, n <= 32, 0 <= K < m * n), the number of rows, column and holes. In the next k lines, there is a pair of integers (x, y) in each line, which represents a hole in the y-th row, the x-th column.

Output

If the board can be covered, output "YES". Otherwise, output "NO".

Sample Input

4 3 2
2 1
3 3

Sample Output

YES

Hint

 
A possible solution for the sample input.

Source

POJ Monthly,charlescpp

和 hdu 1507类似，构无向图然后判断匹配数是否等于合法的格数。

心算32*32错了= = RE了两次，开始以为32*32是90+，第二次以为是900+，笔算后才知道是1024..

 //224K    125MS    C++    1731B    2014-06-10 12:44:41
 #include<iostream>
 #include<vector>
 #define N 1050
 using namespace std;
 vector<int>V[N];
 int match[N];
 int vis[N];
 int g[][];
 int dfs(int u)
 {
     for(int i=;i<V[u].size();i++){
         int v=V[u][i];
         if(!vis[v]){
             vis[v]=;
             if(match[v]==- ||  dfs(match[v])){
                 match[v]=u;
                 return ;
             }
         }
     }
     return ;
 }
 int hungary(int n)
 {
     int ret=;
     memset(match,-,sizeof(match));
     for(int i=;i<=n;i++){
         memset(vis,,sizeof(vis));
         ret+=dfs(i);
     }
     return ret;
 }
 int main(void)
 {
     int n,m,k,x,y;
     while(scanf("%d%d%d",&n,&m,&k)!=EOF)
     {
         memset(g,,sizeof(g));
         for(int i=;i<N;i++) V[i].clear();
         for(int i=;i<k;i++){
             scanf("%d%d",&y,&x);
             g[x-][y]=;
         }
         int map[N]={},pos=;
         for(int i=;i<n;i++)
             for(int j=;j<=m;j++)
                 if(!g[i][j]){
                     if(!map[i*m+j]) map[i*m+j]=++pos;
                     int u=map[i*m+j];
                     if(j<m && !g[i][j+]){
                         if(!map[i*m+j+]) map[i*m+j+]=++pos;
                         V[u].push_back(map[i*m+j+]);
                         V[map[i*m+j+]].push_back(u);
                     }
                     if(i<n- &&  !g[i+][j]){
                         if(!map[(i+)*m+j]) map[(i+)*m+j]=++pos;
                         V[u].push_back(map[(i+)*m+j]);
                         V[map[(i+)*m+j]].push_back(u);
                     }
                 }
         //printf("%d\n",pos);
         if(hungary(pos)==pos) puts("YES");
         else puts("NO");
     }
     return ;
 }